Answer:
SEE THE PROCEDURE PLEASE
Step-by-step explanation:
1.
a. The plot is attached below
b. The area is given by
[tex]A=\frac{1}{2}\int_\alpha^\beta [r(\theta)]^2d\theta\\\\A=\frac{1}{2}\int_{0}^{2\pi}[3+2cos\theta]^2d\theta\\\\A=\frac{1}{2}\int_{0}^{2\pi}[9+12cos\theta+4cos^2\theta]d\theta\\\\A=\frac{1}{2}[9(2\pi)+12sin(2\pi)+2(2\pi)+sin(4\pi)]\\\\A=11\pi[/tex]
c.
[tex]L=\int_0^{2\pi}\sqrt{r^2+(\frac{dr}{d\theta})^2}d\theta\\\\L=\int_0^{2\pi}\sqrt{(3+2cos\theta)^2+(-2sin\theta)^2}d\theta=\int_0^{2\pi}\sqrt{13+12cos\theta}d\theta[/tex]
2.
a. The plot is attached below
b. by symmetry:
[tex]A=6*\frac{1}{2}\int_{-\frac{\pi}{6}}^{\frac{pi}{6}}16cos^2\theta d\theta\\\\A=6[4\sqrt{3}+\frac{8\pi}{3}]=24\sqrt{3}+16\pi[/tex]
c.
[tex]L=6*\int_{-\frac{\pi}{6}}^{\frac{\pi}{6}}\sqrt{16cos^2\theta-144sin^23\theta} d\theta[/tex]
HOPE THIS HELPS!!
